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So, start with the set [0,1] of the real line. Remove the middle third, and keep removing the middle thirds of the remaining line segments as usual when making the Cantor set.

Each time you remove a 3rd, add the middle point of the line segment you just removed to another set, S (initially empty). In the end you have the Cantor set C, and another set, S. Are the points in S countable?

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One obvious way of checking that $S$ is countable is to use that $\mathbb Q$ itself is countable, and to notice that $S\subset \mathbb Q$. – Andrés E. Caicedo Jan 29 '13 at 4:00
up vote 3 down vote accepted

Yes, $S$ is countable. At stage $n$ you removed $2^n$ open intervals, so you added $2^n$ points to $S$. $S$ is therefore the union of countably infinitely many finite sets and is therefore countable.

Another way to prove this is to notice that the open intervals that are removed to form the middle-thirds Cantor set are pairwise disjoint, and each contains exactly one point of $S$. Let $\mathscr{I}$ be the set of removed intervals. For each $s\in S$ let $I_s$ be the member of $\mathscr{I}$ containing $s$, and let $q_s\in I_s\cap\Bbb Q$. Clearly the map from $S\to\Bbb Q:s\mapsto q_s$ is injective (one-to-one), so $|S|\le|\Bbb Q|$, and $S$ is therefore countable.

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placeholder text i accidentally pressed enter – God Of Djinns Jan 29 '13 at 3:35
OK but I'm still confused about something. What happens to the points in S as the number of iterations approaches infinity? In other words, if I said that the "first" element of S maps to 1/2, and the next 2 to 1/4 and 3/4, etc. what would happen if I tried to map, say pi/4 to a point in S by following this pattern, going left if that element of S was greater than pi/4 or right otherwise? I would presumably either not be able to complete this or end up on a point in C, right? – God Of Djinns Jan 29 '13 at 3:40
@GodOfDjinns: Actually, the second and third centres of deleted intervals are $1/6$ and $5/6$. Or are you now mapping the points of $S$ to some different subset of $[0,1]$? – Brian M. Scott Jan 29 '13 at 3:43
You can visualize it as a binary tree with 1/2 on the first level, 1/4 and 3/4 on the second, k/2^n for all k in {1,...,2^n-1} on the nth level/iteration etc. So I'm mapping between points in S and number in R that way. Sorry if that's unclear. Also I'm sure I'm just missing something really obvious here. – God Of Djinns Jan 29 '13 at 3:55
@GodOfDjinns: You get precisely the dyadic rationals in $(0,1)$, i.e., the rational numbers that can be expressed in lowest terms in the form $\frac{k}{2^n}$ for some odd positive integer $k$ and positive integer $n$. This is a countable dense subset of $(0,1)$. – Brian M. Scott Jan 29 '13 at 3:59

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